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Maths
A Level

The numbers a, b, c and d satisfy the following equations: a + b + 3c + 4d = k; 5a = 3b = 2c = d. What is the smallest value for k for which a, b, c and d are all positive integers

  1. 5a = 3b = 2c = d. d must be a multiple of 5, 3 and 2, therefore the smallest possible value for d is 30. This sets a = 6, b = 10 and c = 152) a + b + 3c + 4d = 6 + 10 + 3x15 + 4x30 = 181 k = 181
    Answered by Michael H. Maths tutor
    3807 Views

How can I remember when a turning point of a function is a maximum or a minimum?

The key is to look at the first and second derivatives of that function. Remember that a turning point always has the first derivative equal to zero. Then, the sign of the second derivative indicates if t...

Answered by Titus D. Maths tutor
5967 Views

Using the substitution x = 2cosu, find the integral of dx/((x^2)(4-x^2)^1/2), evaluated between x=1 and x=sqrt(2).

Starting with x=2cosu, rearrange for u to get u=arccos(x/2), then find the upper and lower limits of the integral. We find that our lower limit goes from 1 to pi/3, ad our upper limit goes from root 2 to ...

Answered by Tayn D. Maths tutor
6836 Views

Calculate the integral of e^x*sin x

1/2 *e^(x)(sin x -cos x)

Answered by Maths tutor
2757 Views

Find the equation of the tangent to curve y=5x^2-2x+3 at the point x=0

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